How many grams of Al2O3 are formed from 15L of O2 at 97.5kPa and at 21degree celcius.
To determine the number of grams of Al2O3 formed from 15L of O2, you'll need to use the ideal gas law and stoichiometry.
1. Start by converting the given temperature from Celsius to Kelvin. The equation is:
Kelvin = Celsius + 273.15
So, the temperature is: 21 + 273.15 = 294.15K
2. Next, use the ideal gas law equation, which is:
PV = nRT
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
Convert the pressure from kilopascals (kPa) to atmospheres (atm) by dividing by 101.325:
97.5 kPa ÷ 101.325 kPa/atm = 0.961 atm
3. Rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substitute the values into the formula:
n = (0.961 atm)(15 L) / (0.0821 L.atm/mol.K)(294.15 K)
Calculate the value:
n = 0.477 moles of O2
4. The balanced chemical equation for the formation of Al2O3 is:
4Al + 3O2 → 2Al2O3
From the balanced equation, we know that 3 moles of O2 react to form 2 moles of Al2O3.
5. Use the mole ratio to determine the number of moles of Al2O3:
(0.477 moles O2) × (2 moles Al2O3 / 3 moles O2) = 0.318 moles Al2O3
6. Finally, calculate the molar mass of Al2O3:
Al = 26.98 g/mol
O = 16.00 g/mol (There are 3 oxygen atoms in Al2O3, so the molar mass of O2 is 3 x 16 g/mol)
Molar mass of Al2O3 = (2 x 26.98 g/mol) + (3 x 16.00 g/mol) = 101.96 g/mol
7. Calculate the number of grams using the moles of Al2O3 and its molar mass:
grams = moles × molar mass
grams = 0.318 moles Al2O3 × 101.96 g/mol Al2O3
Calculate the value:
grams = 32.47 grams of Al2O3
Therefore, approximately 32.47 grams of Al2O3 are formed from 15L of O2 at 97.5kPa and 21 degrees Celsius.